Sn+an=(n-1)/n(n+1),求an

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Sn+an=(n-1)/n(n+1),求an
Sn+an=(n-1)/n(n+1),求an

Sn+an=(n-1)/n(n+1),求an
n=1时,S1+a1=2a1=(1-1)/(1×2)=0 a1=0
n≥2时,
Sn+an=(n-1)/[n(n+1)]=n/[n(n+1)]-1/[n(n+1)]=1/(n+1)-[1/n -1/(n+1)]=2/(n+1) -1/n
S(n-1)+a(n-1)=2/n -1/(n-1)
Sn+an-S(n-1)-a(n-1)=2an-a(n-1)=2/(n+1)-1/n-2/n +1/(n-1)=[2/(n+1)-2/n]-[1/n -1/(n-1)]
2an+[2/n-2/(n+1)]=a(n-1)+[1/(n-1) -1/n]
[an+1/n -1/(n+1)]/[a(n-1)+1/(n-1)-1/n]=1/2,为定值.
a1+1/1-1/2=0+1-1/2=1/2
数列{an +1/n -1/(n+1)}是以1/2为首项,1/2为公比的
.
an +1/n -1/(n+1)=1/2ⁿ
an=1/2ⁿ -1/n +1/(n+1)
n=1时,a1=1/2 -1/1+1/2=0,同样满足.
数列{an}的
为an=1/2ⁿ -1/n +1/(n+1).